Random Signal Notes

Signals

We’ll first import the following libraries:

import numpy as np
import matplotlib.pyplot as plt
from IPython.display import Audio

Suppose we have the following signal which is sampled at 50 Hz and has a duration of 2 seconds.

fs = 50
duration = 2
freq = 2
N = np.arange(int(fs * duration))
t = N / fs       
y = np.cos(2 * np.pi * freq * t)


plt.plot(t, y)
plt.xlabel("Seconds")
plt.ylabel("Amplitude")
plt.title("Wave Example")

plt.grid(True)

Figure 1: Signal example

The signal above has a frequency of 2 Hz (fs) which means that it completes 2 cycles in 1 second.

Definition: Period

The period of the signal is how long it takes for a signal to complete 1 cycle by taking

\[ t_0 = \frac{1}{f_s} \]

where \(f_s\) is the frequency of the signal.

In Figure 1, the signal has a period of \(t_0 = \frac{1}{f_s} = \frac{1}{2} = 0.5\) seconds.

Notes on Music Notes

The human ear can hear frequencies between 20 Hz and 20 kHz. The sampling rate is 44.1 kHZ because of something called the Nyquist–Shannon sampling theorem which states that the sampling rate has to be at least twice of the maximum frequency of the signal which indeed \(2 \cdot 20 \text{ kHz} < 44.1 \text{ kHZ}\). There’s also more history for why the sampling rate is 44.1 kHZ here.

Now, as for the frequencies of musical notes in Western music, we can use the formula below:

\[ f = 2^{(n/12)} * 440 \]

The formula represents the frequency of the music note that is \(n\) semitones away from A4 (440 Hz). Each increasing semitone step increases the frequency by the ratio of \(2^{(n/12)}\).

Examples

\(n = 0 \rightarrow 2^{(0/12)} * 440 = 440 \text{ Hz} = A4\)
\(n = 1 \rightarrow 2^{(1/12)} * 440 \approx 466.163 \text{ Hz} = A \sharp 4 / B\flat 4\)
\(n = 12 \rightarrow 2^{(12/12)} * 440 = 880 \text{ Hz} = A5\)
\(n = -12 \rightarrow 2^{(-12/12)} * 440 = 220 \text{ Hz} = A3\)

Notice that doubling the frequency represents an octave increase of a music note. Now, if we assume that 20 kHz is the maximum frequency that the human ear can hear, then the highest named musical note can be found by solving for \(n\) when \(f = 20 \text{ kHz}\).

\[ \begin{align*} 20000 &= 2^{\frac{n}{12}} \cdot 440 \\ \frac{20000}{440} &= 2^{\frac{n}{12}} \\ 45.45 &= 2^{\frac{n}{12}} \\ \log_2(45.45) &= \frac{n}{12} \\ n &= 12 \cdot \log_2(45.45) \\ n &\approx 66.0745 \end{align*} \]

When \(n \approx 66\), this is 66 semitones aboves A4 and corresponds to the note \(D \sharp 10/ E\flat 10\).

Let’s take a listen of the music notes, starting from A3 to D#10!

fs = 44100
note_duration = 0.5
pause_duration = 0.10

N = np.arange(int(fs * note_duration))
t = N / fs

pause_arr = np.zeros(int(fs * pause_duration))

def get_frequency(n):
    """
    Return the frequency of the note n semitones starting from A4 (440 Hz).
    """
    return 2**(n / 12) * 440



audio_sequence = np.array([])
for n in range(-12, 67):
  freq = get_frequency(n)
  signal_arr = np.cos(2 * np.pi * freq * t)
  audio_sequence = np.concatenate((audio_sequence, signal_arr, pause_arr))

Audio(data=audio_sequence, rate=fs)